a, |x - 3| - 5 = 7x
=> |x - 3| = 7x + 5
Đk: 7x + 5 ≥ 0 => x ≥ -5/7
Ta có: |x - 3| = 7x + 5
\(\Rightarrow\orbr{\begin{cases}x-3=7x+5\\x-3=-7x-5\end{cases}\Rightarrow}\orbr{\begin{cases}-6x=8\\8x=-2\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{-4}{3}\left(ktm\right)\\x=\frac{-1}{4}\left(tm\right)\end{cases}}\Rightarrow x=\frac{-1}{4}\)
b, 209 - |x - 209| = x
=> |x - 209| = 209 - x
Đk: 209 - x ≥ 0 => x ≤ 209
Ta có: |x - 209| = 209 - x
\(\Rightarrow\orbr{\begin{cases}x-209=209-x\\x-209=x-209\end{cases}\Rightarrow}\orbr{\begin{cases}2x=418\\0x=0\forall x\le209\end{cases}\Rightarrow\orbr{\begin{cases}x=209\\x\le209\end{cases}}}\)
=> x ≤ 209
c, (x - 1)2008 + (y - 1)2008 + |x + y + z| = 0
Vì (x - 1)2008 ≥ 0 ; (y - 1)2008 ≥ 0 ; |x + y + z| ≥ 0
=> (x - 1)2008 + (y - 1)2008 + |x + y + z| ≥ 0
Dấu " = " xảy ra <=> \(\hept{\begin{cases}\left(x-1\right)^{2008}=0\\\left(y-1\right)^{2008}=0\\\left|x+y+z\right|=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x-1=0\\y-1=0\\x+y+z=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=1\\1+1+z=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=y=1\\z=-2\end{cases}}}\)
